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$\int^{b}_{a} x \text{ dx}$?

I agree it forms a triangle, but in the case where $b <0$, there's another triangle we have to factor in.

Is there a formula takes takes both of these into account, and just spits out the answer?

For ex) $\int^{2}_{-1} x \text { dx}$

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    Note (or, better, prove) that$\int^{2}_{-1} x \,dx = \int^{2}_{1} x \,dx\,$. Hint: $\int^{a}_{-a} x \,dx = 0\,$ for $\forall a\,$. – 2017-01-17
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    yes. $\frac{b^2 - a^2}{2}$. I assume you're in Calculus. Stay tuned! $x^2/2$ is an antiderivative for $x$, and then the fundamental theorem of calculus gives you the result. – 2017-01-17
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    Yeah @Callus, I still gotta evaluate using area instead of antiderivative, if you put your answer in the description I can mark as correct – 2017-01-17
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    @dxiv, wow. I didn't know that ... thanks, I wonder why it works .. – 2017-01-17
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    @KSplitX $\int^{a}_{-a} f(x) \,dx = 0\,$ for any function that is odd i.e. $f(-x)=-f(x)\,$. Think at the point symmetry with respect to the origin, the areas left and right of the origin are congruent, but have opposite signs and cancel out. – 2017-01-17
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    Remember that if the area is under the $x$-axis, it counts as negative. – 2017-01-17

4 Answers 4

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The integral from $0$ to $b$ is $\frac12b^2$:

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The integral from $0$ to $a$ is $\frac12a^2$:

enter image description here

Your integral is then as follows:

$$\int_a^bx\ dx=\int_0^bx\ dx-\int_0^ax\ dx=\frac12b^2-\frac12a^2$$

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    Nice application of desmos xD – 2017-01-17
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    @KSplitX Didn't quite know how to make the $a,b$ fit in better though :-P – 2017-01-17
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    Although in the integral you wrote $x^2$? I think you should fix – 2017-01-17
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    @KSplitX >.> What $x^2$? – 2017-01-17
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    LOOL nice edit :P – 2017-01-17
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The formula is

$$ \frac{1}{2} \left( b^2-a^2 \right) $$

Do you know why? What is the indefinite integral of $x$?

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Yes it is a standard integration.

$$ \int _a^b x \; dx,$$ where upon integration becomes $$ \frac{x^2}{2} |^b_a = \frac{1}{2}(b^2 - a^2).$$ Voila! Enjoy.

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Assuming $a<0$ and $b>0$. Spit the integral into, (this is the same as breaking up signed areas into two)

$$\int_{a}^{0} x dx+\int_{0}^{b} x dx$$

The first is going to be negative as it is below the $y$ axis, but it will in magnitude represent a triangle with base $a$ and height $a$. The second is going to be positive and represent a triangle with base $b$ and height $b$. So the integral is,

$$-\frac{1}{2}(a)(a)+\frac{1}{2}(b)(b)=\frac{1}{2}b^2-\frac{1}{2}a^2$$

There are a couple more cases to check $a,b>0$ which results in a trapezoid with base $b-a$ and average height $\frac{b+a}{2}$. So again we get $(b-a)\frac{b+a}{2}=\frac{1}{2}(b^2-a^2)$. Then the case $a,b<0$ is going to be similar to this. And the case $a>0, b<0$ is going to be similar to the very first case we considered.